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0=(2y-1)(8-y) Let y=u and y=d be unique solutions to the given equation. What is the value of u*d ?

0=(2y−1)(8−y) 0=(2 y-1)(8-y) \newlineLet y=u y=u and y=d y=d be unique solutions to the given equation. What is the value of u⋅d u \cdot d ?

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Q. 0=(2y−1)(8−y) 0=(2 y-1)(8-y) \newlineLet y=u y=u and y=d y=d be unique solutions to the given equation. What is the value of u⋅d u \cdot d ?
  1. Set First Factor Equal: To find the unique solutions to the equation (2y−1)(8−y)=0(2y-1)(8-y) = 0, we need to set each factor equal to zero and solve for yy. First, let's set the first factor equal to zero: 2y−1=02y - 1 = 0. Adding 11 to both sides gives us 2y=12y = 1. Dividing both sides by 22 gives us y=12y = \frac{1}{2}.
  2. Solve for y: Now, let's set the second factor equal to zero: 8−y=08 - y = 0. Subtracting 88 from both sides gives us −y=−8-y = -8. Multiplying both sides by −1-1 gives us y=8y = 8.
  3. Set Second Factor Equal: We have found the two unique solutions to the equation: y=12y = \frac{1}{2} and y=8y = 8. To find the product of these solutions, we multiply them together: (12)×8\left(\frac{1}{2}\right) \times 8. This simplifies to 44.

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